<mrdarrett@gmail.com> wrote in message
news:5e56fa73-9e09-4ce6-bc2f-7b6ba1dd1ec8@googlegroups.com...
> On Wednesday, June 12, 2013 10:03:26 AM UTC-7, Jim Thompson wrote:
>
> ...
>
>> >Well, sure, that adds $thousands to both sides of the equation... but
>> >you add them whether or not you use batteries XD
>>
>> >
>>
>> >One thing I do not know... how much power will a typical wall-wart
>> >consume, radiating heat, just from being plugged in? That, I do not
>> >know. When I get a spare moment I'll calculate the breakeven heat loss
>> >power cost ;)
>>
>> >
>>
>> >M
>>
>>
>>
>> For low power drain, a wall-wart is probably quite inefficient.
>
>
> Yes... looks like if the wall wart wastes more than 0.2 watts, it will
> cost more than the battery... hmm...
You could always go for the "wattless dropper", its a little dodgy insofar
as it puts the current waveform ahead of the voltage and cons the
electricity meter a bit.
There are a few gotchas to watch out for - especially for a very low power
use like a clock.
Around the late 70's; the UK TV maker TCE adopted the "wattless dropper" for
the 300mA series heater chain, this consisted of a 4.3uF capacitor in series
instead of a dropper resistor, for 50Hz applications the capacitance is
directly scalable for different current values - but don't forget to factor
in the reduced Xc for a 60Hz supply.
You need a significant resistor in series with the capacitor to absorb turn
on surge and any mains borne spikes, its basically a constant current supply
so you must use shunt regulation for intermittent loads (like an electronic
clock escapement).
The simplest configuration is the capacitor in series with the input to a
bridge rectifier - but then the DC rails swing +/- of the ground potential,
a better approach is to use the charge-pump style voltage doubler
configuration - at least then you have a ground referenced rail.
Also, don't forget a mains input bleed resistor to discharge the capacitor
when you pull the plug out the socket - needless to say, this resistor must
be rated to take continuous mains.
Reply by ●June 12, 20132013-06-12
On Wednesday, June 12, 2013 10:03:26 AM UTC-7, Jim Thompson wrote:
...
> >Well, sure, that adds $thousands to both sides of the equation... but you add them whether or not you use batteries XD
>
> >
>
> >One thing I do not know... how much power will a typical wall-wart consume, radiating heat, just from being plugged in? That, I do not know. When I get a spare moment I'll calculate the breakeven heat loss power cost ;)
>
> >
>
> >M
>
>
>
> For low power drain, a wall-wart is probably quite inefficient.
Yes... looks like if the wall wart wastes more than 0.2 watts, it will cost more than the battery... hmm...
> But I am toying with powering an outdoor clock via a wall-wart, since
>
> the clock is made of flagstone and is a pain to take down.
Yes! I also was struck with the genius of Phil's suggestion. Quite elegant, and needs few parts.
Michael
Reply by Jim Thompson●June 12, 20132013-06-12
On Wed, 12 Jun 2013 09:54:11 -0700 (PDT), mrdarrett@gmail.com wrote:
>On Wednesday, June 12, 2013 9:46:07 AM UTC-7, Jim Thompson wrote:
>> On Wed, 12 Jun 2013 09:40:43 -0700 (PDT), mrdarrett@gmail.com wrote:
>>
>>
>>
>> >On Tuesday, May 28, 2013 1:08:34 AM UTC-7, Phil Allison wrote:
>>
>> >> <videoman@ccountry.net>
>>
>> >>
>>
>> >>
>>
>> >>
>>
>> >>
>>
>> >>
>>
>> >> Yeah, the clock operates on one AA alkaline battery for about 1 year at 1.5
>>
>> >>
>>
>> >> volts.
>>
>> >>
>>
>> >>
>>
>> >>
>>
>> >>
>>
>> >>
>>
>> >> ** Changing to a wall wart will cost much more in electricity than using AA
>>
>> >>
>>
>> >> cells does.
>>
>> >
>>
>> >
>>
>> >It will???
>>
>> >
>>
>> >Here (California) we can get a 4-pack of AA batteries at the 99 Cent store ($1.08 after sales taxes). So, $0.27 per year.
>>
>> >
>>
>> >Running 3 volts at 1000 ohms (I assume two 500 ohm resistors to provide a little extra juice to charge the capacitor) gives 0.009 watts (P = V^2/R).
>>
>> >
>>
>> >1 yr x (365 d/yr) x (24 h/d) x ($0.15/kw-hr) x (1 kw/1000 w) x 0.009 W gives... $0.012/yr.
>>
>> >
>>
>> >$0.27 >> $0.012
>>
>>
>>
>> Plus Californica tax >:-}
>
>
>Well, sure, that adds $thousands to both sides of the equation... but you add them whether or not you use batteries XD
>
>One thing I do not know... how much power will a typical wall-wart consume, radiating heat, just from being plugged in? That, I do not know. When I get a spare moment I'll calculate the breakeven heat loss power cost ;)
>
>M
For low power drain, a wall-wart is probably quite inefficient.
But I am toying with powering an outdoor clock via a wall-wart, since
the clock is made of flagstone and is a pain to take down.
...Jim Thompson
--
| James E.Thompson | mens |
| Analog Innovations | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| San Tan Valley, AZ 85140 Skype: Contacts Only | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |
I love to cook with wine. Sometimes I even put it in the food.
Reply by ●June 12, 20132013-06-12
On Wednesday, June 12, 2013 9:46:07 AM UTC-7, Jim Thompson wrote:
> On Wed, 12 Jun 2013 09:40:43 -0700 (PDT), mrdarrett@gmail.com wrote:
>
>
>
> >On Tuesday, May 28, 2013 1:08:34 AM UTC-7, Phil Allison wrote:
>
> >> <videoman@ccountry.net>
>
> >>
>
> >>
>
> >>
>
> >>
>
> >>
>
> >> Yeah, the clock operates on one AA alkaline battery for about 1 year at 1.5
>
> >>
>
> >> volts.
>
> >>
>
> >>
>
> >>
>
> >>
>
> >>
>
> >> ** Changing to a wall wart will cost much more in electricity than using AA
>
> >>
>
> >> cells does.
>
> >
>
> >
>
> >It will???
>
> >
>
> >Here (California) we can get a 4-pack of AA batteries at the 99 Cent store ($1.08 after sales taxes). So, $0.27 per year.
>
> >
>
> >Running 3 volts at 1000 ohms (I assume two 500 ohm resistors to provide a little extra juice to charge the capacitor) gives 0.009 watts (P = V^2/R).
>
> >
>
> >1 yr x (365 d/yr) x (24 h/d) x ($0.15/kw-hr) x (1 kw/1000 w) x 0.009 W gives... $0.012/yr.
>
> >
>
> >$0.27 >> $0.012
>
>
>
> Plus Californica tax >:-}
Well, sure, that adds $thousands to both sides of the equation... but you add them whether or not you use batteries XD
One thing I do not know... how much power will a typical wall-wart consume, radiating heat, just from being plugged in? That, I do not know. When I get a spare moment I'll calculate the breakeven heat loss power cost ;)
M
Reply by Jim Thompson●June 12, 20132013-06-12
On Wed, 12 Jun 2013 09:40:43 -0700 (PDT), mrdarrett@gmail.com wrote:
>On Tuesday, May 28, 2013 1:08:34 AM UTC-7, Phil Allison wrote:
>> <videoman@ccountry.net>
>>
>>
>>
>>
>>
>> Yeah, the clock operates on one AA alkaline battery for about 1 year at 1.5
>>
>> volts.
>>
>>
>>
>>
>>
>> ** Changing to a wall wart will cost much more in electricity than using AA
>>
>> cells does.
>
>
>It will???
>
>Here (California) we can get a 4-pack of AA batteries at the 99 Cent store ($1.08 after sales taxes). So, $0.27 per year.
>
>Running 3 volts at 1000 ohms (I assume two 500 ohm resistors to provide a little extra juice to charge the capacitor) gives 0.009 watts (P = V^2/R).
>
>1 yr x (365 d/yr) x (24 h/d) x ($0.15/kw-hr) x (1 kw/1000 w) x 0.009 W gives... $0.012/yr.
>
>$0.27 >> $0.012
Plus Californica tax >:-}
>
>
>>
>>
>>
>> PLUS if the AC power ever goes off for a while, the clock will stop and
>>
>> thereafter show the wrong time.
>
>
>No argument there
>
>M
...Jim Thompson
--
| James E.Thompson | mens |
| Analog Innovations | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| San Tan Valley, AZ 85140 Skype: Contacts Only | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |
I love to cook with wine. Sometimes I even put it in the food.
Reply by ●June 12, 20132013-06-12
On Tuesday, May 28, 2013 1:08:34 AM UTC-7, Phil Allison wrote:
> <videoman@ccountry.net>
>
>
>
>
>
> Yeah, the clock operates on one AA alkaline battery for about 1 year at 1.5
>
> volts.
>
>
>
>
>
> ** Changing to a wall wart will cost much more in electricity than using AA
>
> cells does.
It will???
Here (California) we can get a 4-pack of AA batteries at the 99 Cent store ($1.08 after sales taxes). So, $0.27 per year.
Running 3 volts at 1000 ohms (I assume two 500 ohm resistors to provide a little extra juice to charge the capacitor) gives 0.009 watts (P = V^2/R).
1 yr x (365 d/yr) x (24 h/d) x ($0.15/kw-hr) x (1 kw/1000 w) x 0.009 W gives... $0.012/yr.
$0.27 >> $0.012
>
>
>
> PLUS if the AC power ever goes off for a while, the clock will stop and
>
> thereafter show the wrong time.
No argument there
M
Reply by Phil Allison●June 3, 20132013-06-03
"David Eather"
> He could just use 2 diodes in series for the shunt regulator
** The voltage is too low at 1.1V to 1.3V to reliably operate a 1.5V clock -
often they will not run on a NiCd or NiMH cell.
Three 1 amp diodes and a 330 ohm resistor might be the go.
... Phil
Reply by Ian Field●June 2, 20132013-06-02
"David Eather" <eather@tpg.com.au> wrote in message
news:op.wx2ljfpswei6gd@phenom-pc...
> On Sun, 02 Jun 2013 06:47:38 +1000, Ian Field
> <gangprobing.alien@ntlworld.com> wrote:
>
>>
>>
>> "Daniel" <videoman@ccountry.net> wrote in message
>> news:2debb754-eb4d-4f7c-bd80-bc5024b38b87@googlegroups.com...
>>> Hi, I have recently been trying to get 1.5 volts (or anything other
>>> than the power supply's rated 3 volts for that matter) out of a
>>> wall-wart 3 volt power supply. The wall-wart power supply is dc in
>>> output at 3 volts and even if I connect a 300 ohm resistor in series
>>> with the wall-wart power supply and the load (a 1.5 volt clock, old
>>> style analog clock)and when i measure the voltage with the voltmeter in
>>> place of the load (but remember the resistor is still in series) I keep
>>> getting 3 volts dc measured. I even put a 20 ohm resistor in parallel
>>> with the 300 ohm resister and still no change in measurement. I bought
>>> a older style wall-wart with a 1.5 volt setting and it measures at 3
>>> volts
>>
>> What you need is an adjustable shunt regulator + dropper resistor - the
>> older TL431 only went down to 2.45V, but there are newer versions with
>> 1.25V reference:
>>
>> http://www.nxp.com/documents/leaflet/75017148.pdf
>>
>> As someone else mentioned - if the power goes out your clock will show
>> the wrong time, never tried it, but you might get away with backing it
>> up with a supercapacitor if the current draw is really that low.
>
> He could just use 2 diodes in series for the shunt regulator or a
> transistor and 2 resistors if he wants to get fancy (and a cap for the
> current peak when the clock 'ticks')
>
A common trick used with the MK484 (ZN414) AM radio, but diodes have a
fairly soft knee compared to a programmable zener of the TL431 general type.
Reply by David Eather●June 2, 20132013-06-02
On Sun, 02 Jun 2013 06:47:38 +1000, Ian Field
<gangprobing.alien@ntlworld.com> wrote:
>
>
> "Daniel" <videoman@ccountry.net> wrote in message
> news:2debb754-eb4d-4f7c-bd80-bc5024b38b87@googlegroups.com...
>> Hi, I have recently been trying to get 1.5 volts (or anything other
>> than the power supply's rated 3 volts for that matter) out of a
>> wall-wart 3 volt power supply. The wall-wart power supply is dc in
>> output at 3 volts and even if I connect a 300 ohm resistor in series
>> with the wall-wart power supply and the load (a 1.5 volt clock, old
>> style analog clock)and when i measure the voltage with the voltmeter in
>> place of the load (but remember the resistor is still in series) I keep
>> getting 3 volts dc measured. I even put a 20 ohm resistor in parallel
>> with the 300 ohm resister and still no change in measurement. I bought
>> a older style wall-wart with a 1.5 volt setting and it measures at 3
>> volts
>
> What you need is an adjustable shunt regulator + dropper resistor - the
> older TL431 only went down to 2.45V, but there are newer versions with
> 1.25V reference:
>
> http://www.nxp.com/documents/leaflet/75017148.pdf
>
> As someone else mentioned - if the power goes out your clock will show
> the wrong time, never tried it, but you might get away with backing it
> up with a supercapacitor if the current draw is really that low.
He could just use 2 diodes in series for the shunt regulator or a
transistor and 2 resistors if he wants to get fancy (and a cap for the
current peak when the clock 'ticks')
Reply by Ian Field●June 1, 20132013-06-01
"Daniel" <videoman@ccountry.net> wrote in message
news:2debb754-eb4d-4f7c-bd80-bc5024b38b87@googlegroups.com...
> Hi, I have recently been trying to get 1.5 volts (or anything other than
> the power supply's rated 3 volts for that matter) out of a wall-wart 3
> volt power supply. The wall-wart power supply is dc in output at 3 volts
> and even if I connect a 300 ohm resistor in series with the wall-wart
> power supply and the load (a 1.5 volt clock, old style analog clock)and
> when i measure the voltage with the voltmeter in place of the load (but
> remember the resistor is still in series) I keep getting 3 volts dc
> measured. I even put a 20 ohm resistor in parallel with the 300 ohm
> resister and still no change in measurement. I bought a older style
> wall-wart with a 1.5 volt setting and it measures at 3 volts
What you need is an adjustable shunt regulator + dropper resistor - the
older TL431 only went down to 2.45V, but there are newer versions with 1.25V
reference:
http://www.nxp.com/documents/leaflet/75017148.pdf
As someone else mentioned - if the power goes out your clock will show the
wrong time, never tried it, but you might get away with backing it up with a
supercapacitor if the current draw is really that low.